ON THE COMPOSITUM OF ALL DEGREE d EXTENSIONS OF A NUMBER FIELD. Itamar Gal and Robert Grizzard

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1 ON THE COMPOSITUM OF ALL DEGREE d EXTENSIONS OF A NUMBER FIELD Itamar Gal and Robert Grizzard Department of Mathematics, The University of Texas at Austin April 22, 2013 Abstract. Let k be a number field, and denote by k [d] the compositum of all degree d extensions of k in a fixed algebraic closure. We first consider the question of whether all subextensions of degree less than d lie in k [d]. We show that this occurs if and only if d < 5. Secondly, we consider the question of whether there exists a constant c such that if K/k is a finite subextension of k [d], then K is generated over k by elements of degree at most c. We show that such a constant exists if and only if d < 3. This question becomes more interesting when one restricts attention to Galois extensions K/k. In this setting, we show that such a constant does not exist for d even or non-squarefree. If d is prime, we prove that all finite Galois subextensions of k [d] are generated over k by elements of degree at most d. 1. Introduction Let k be a field. Throughout this paper, all extensions of k will be assumed to lie in a fixed algebraic closure k. We are interested in fields obtained by adjoining to k all roots of irreducible polynomials of a given degree d. For any positive integer d we will write k [d] = k(β (1) [k(β) : k] = d), and (2) k (d) = k(β [k(β) : k] d) = k [2] k [3] k [4] k [d]. We have k [1] = k (1) = k, and when d > 1 it is clear that k [d] and k (d) are normal extensions of k. We are primarily interested in the case where k is a number field, in which case these are infinite Galois extensions. When d > 2 it is natural to ask what polynomials of degree less than d split in k [d]. If c < d and all irreducible polynomials of degree c split in k [d], then k [c] k [d]. Notice that this occurs in particular when c divides d, since every degree c extension admits a degree d/c extension. If all polynomials of degree less than d split in k [d], then k [d] = k (d). We will prove the following results along these lines. Theorem 1. If k is a number field, then (a) k [2] k [d] for all d 2, (b) k [3] k [4], and (c) for each d 5, there exists a prime p < d such that k [p] k [d] Mathematics Subject Classification. 12F10, 11R21, 20B05. Key words and phrases. number fields, infinite algebraic extensions, Galois theory, permutation groups. The author s research was partially supported by a grant from the National Security Agency, H Many of our results contain the hypothesis that k is a number field, or similar. However, the astute reader will notice after reading the proofs that this hypothesis could be replaced with more technical restrictions on the field k specifically, that certain embedding problems have solutions over k. 1

2 2 ON THE COMPOSITUM Parts (a) and (b) follow from some well-known solutions to simple embedding problems over number fields. We give original proofs of some of these results, but parts (a) and (b) are fundamentally restatements of established facts. Our main contribution in this theorem is part (c). The following corollary is immediate. Corollary 1. If k is a number field, then k [d] = k (d) if and only if d < 5. We now introduce the notion of boundedness for an extension of fields. We will use this language to state our remaining results. Definition 1. We say an infinite extension M of k is bounded over k (or that M/k is bounded) if there exists a constant c such that all finite subextensions of M/k can be generated by elements of degree less than or equal to c. If there is no such c, we say that M/k is unbounded. If all finite Galois subextensions of M/k can be generated by elements of degree less than or equal to c, we say M/k is Galois bounded; otherwise we say M/k is Galois unbounded. It was first shown in [7] that, for a number field k, the extension k (d) /k is not in general Galois bounded. We will address the question of how boundedness and Galois boundedness depend on d for the fields k (d) and k [d]. Further restricting attention to abelian Galois extensions greatly simplifies the discussion. It is easily seen that k (d) ab is bounded over k (see [5] for a proof). In the case where k is a number field, E. Bombieri and U. Zannier ask in [3] whether, for any given constant T, only finitely many points in k (d) have absolute Weil height (see [2] for a definition) at most T. Such a finiteness property is called the Northcott property. This problem has been further discussed in [18] and [6], but remains open. In [3] it is proved that this property is enjoyed by k (d) ab, and the boundedness of k(d) ab /k plays a role in the proof. The authors of the present work are hopeful that understanding the boundedness properties in k [d] and k (d) will be useful in understanding these questions. The following theorems summarize our results on boundedness and Galois boundedness. Theorem 2. If k is a number field, then k [d] is bounded over k if and only if d 2. Theorem 3. If k is any field and p is a prime, then k [p] is Galois bounded over k. More precisely, all finite subextensions of k [p] /k can be generated by elements of degree at most p over k. We will establish the following partial converse to Theorem 3. Theorem 4. If k is a number field or global function field and d > 2, then k [d] /k is Galois unbounded in the following cases: (a) d is divisible by a square; (b) d is even; (c) d is divisible by an integer of the form pm, where p is prime and m 1 (mod p). Theorem 3 is an immediate corollary of Theorem 6, a purely group theoretic statement, which we state and prove in section 5. This proof relies on the Classification Theorem for Finite Simple Groups. If K/k is a finite separable extension, then K k [d] if and only if K is contained in some Galois extension L/k, where L is the splitting field of f 1 (x) f n (x), with f i (x) an

3 ON THE COMPOSITUM 3 irreducible separable polynomial of degree d for i {1,..., n}. We call such an extension L a degree d compositum over k. For each i let G i be the Galois group of f i, and let G = Gal(L/k). Each group G i is a transitive group of degree d, i.e. G i acts faithfully and transitively on a set with d elements. Notice that G is a subdirect product of G 1 G n, meaning a subgroup of the direct product whose projection onto each factor is surjective. We will often write G sd G 1 G n to describe such a relation. This observation and its converse are summarized by the following elementary proposition. Proposition 1. Let L/k be a finite Galois extension of fields. The following are equivalent: (a) L is generated by elements of degree d over k; (b) in Gal(L/k) the trivial group is the intersection of subgroups of index d; (c) Gal(L/k) is a subdirect product of transitive groups of degree d. Proof. The equivalence (a) and (b) follows immediately using the Galois correspondence. That (a) implies (c) has already been explained. To see that (c) implies (b), consider point-stabilizers. In view of these observations, we have two goals. First, we wish to understand some key facts about the subgroup structure of subdirect products of transitive groups of a fixed degree d. Second, we must attempt to realize certain group theoretic constructions in the Galois groups of extensions of k. These ideas are discussed in section 2. Acknowledgments The authors would like to thank Daniel Allcock, Sara Checcoli, Joseph Gunther, Andrea Lucchini, Frank Thorne, Jeffrey Vaaler, and anonymous referees for numerous useful communications. We would also like to express our appreciation to the GAP group. Although we did not use any computer calculations directly for any of the results in this paper, we used the GAP software package extensively to improve our understanding of the group theoretic aspect of these problems. 2. Embedding Problems In order to establish Theorem 1, we must discuss the embedding problem in Galois theory. Let K/k be a Galois extension of fields, G a finite group, and N a normal subgroup of G with a short exact sequence (3) 1 N G φ Gal(K/k) 1. This data gives us the embedding problem (K/k, G, N). A solution to the embedding problem is an extension L/k with L K such that Gal(L/k) = G and the natural map Gal(L/k) Gal(K/k) agrees with φ. Hence, a solution to the embedding problem is described by the following commutative diagram. Gal(L/k) (4) φ 1 N G Gal(K/k) 1.

4 4 ON THE COMPOSITUM For our purposes, all that is important is finding an extension L/k such that L K and Gal(L/k) = G, and therefore we will not mention the map φ in what follows. A famous application of this theory is a theorem of Shafarevich, which states that if k any number field or global function field, any solvable group can be realized as the Galois group of some extension of k. Since products of solvable groups are solvable, this allows us to realize a solvable group as the Galois group of infinitely many extensions, whose pairwise intersections are k. A full proof of Shafarevich s Theorem, along with more background on embedding theory, can be found in [15]. If K is a Galois extension of k that is contained in a degree d compositum L over k, we have the standard short exact sequence (5) 1 Gal(L/K) Gal(L/k) Gal(K/k) 1. Thus, using Proposition 1, we have that K k [d] if and only if the following two conditions are met. (i) We can find a group H which is a subdirect product of transitive groups of degree d with some normal subgroup N such that there is a short exact sequence (6) 1 N H Gal(K/k) 1. (ii) We can solve the corresponding embedding problem, i.e. find L K, where L is a degree d compositum over k, such that Gal(L/k) = H. Now let K be the splitting field of an irreducible polynomial of degree m < d in k[x]. In the case m = 2, we must have that Gal(K/k) = C 2, and we have the following. Proposition 2. Let K/k be a quadratic extension of number fields and let d 3. Then there is a solution to the embedding problem (K/k, S d, A d ) arising from (7) 1 A d S d Gal(K/k) 1. This result is due to O. Neumann, and can be found in [16]. This proposition says that, for any integer d 2, any quadratic extension of a number field lies in the splitting field of some polynomial of degree d with symmetric Galois group. This will establish part (a) of Theorem 1, and in particular it tells us that k [3] = k (3). At the end of section 3 we give a very elementary proof of part (a) in the case where k = Q. For part (b) we consider the case m = 3, d = 4. We must have Gal(K/k) = S 3 or C 3, and we will prove the following. Proposition 3. Let k be a number field and let f(x) k[x] be an irreducible cubic with splitting field K. (a) If Gal(K/k) = S 3, then there is a solution to the embedding problem (K/k, S 4, V ) arising from 1 V S 4 Gal(K/k) 1. (b) If Gal(K/k) = C 3, then there is a solution to the embedding problem (K/k, A 4, V ) arising from 1 V A 4 Gal(K/k) 1. This result is well-known, although we have been unable to find an explicit proof in the literature. In [8], for example, the author discusses counting quartic fields with a given cubic resolvent. The main idea of the proof is to find a non-square generator of the cubic

5 ON THE COMPOSITUM 5 field with square norm. Propositions 2 and 3 together imply that k [4] = k (4) for any number field k. To prove part (c) of Theorem 1 we consider the case d 5. We will show that, for certain primes p < d, if Gal(K/k) = C p, then there is no possible subdirect product of transitive subgroups of S d having Gal(K/k) as a quotient. That is, we cannot even find groups H and N satisfying a short exact sequence as in (6) above. In summary, we will see that if d 4, an irreducible polynomial in k[x] of degree less than d splits in the splitting field of a single irreducible polynomial of degree d. When d > 4, however, some irreducible polynomials of degree less than d do not split in any compositum of such splitting fields. We include here a new proof of Proposition 3 using elementary ideas. It is clear that one must find a biquadratic extension L/K (i.e. Gal(L/K) = V ) with certain properties. We will need the following lemma in order to construct this extension. The proof of this lemma can be found at the end of this section. Lemma 1. Let k be a number field, and let f(x) be an irreducible cubic in k[x] with discriminant D and splitting field K. Let F = k( D). There exists a generator α 1 of K/F that is not a square in K, but such that N K/F (α 1 ) is a square in k. This implies that, if α 2 is a Galois conjugate of α 1 over k, then α 1 α 2 is also not a square in K. Proof of Proposition 3. Our goal is to construct a biquadratic extension L/K in such a way as to ensure L/k will be normal and have the correct Galois group (i.e. S 4 or A 4, corresponding respectively to the cases when the Galois group of f(x) is S 3 or C 3 ). By Lemma 1 we can find a generator α 1 of K/F, with Galois conjugates α 2 and α 3, such that α 1 α 2 is not a square in K. (Notice that this means that α 2 α 3, α 1 α 3, α 2, and α 3 are all also not squares in K.) Let β 12 = α 1 α 2, β 13 = α 1 α 3, and β 23 = α 2 α 3 for some choices of the square roots. Notice that {±β 12, ±β 13, ±β 23 } is a complete set of Galois conjugates over k. Let L = K(β 12, β 13 ) (notice that β 23 L). We have that L/K is biquadratic, since β 13 K(β 12 ): if it were, we could write β 13 = a+bβ 12 for some a, b K, b 0; rewriting this as β 13 bβ 12 = a and squaring both sides, we would find that β 23 was in K, which we know to be untrue. To see that L/k is normal, notice that L is the compositum of the normal extensions K/k and k(β 12, β 13 )/k. Let G = Gal(L/k), so either G = 12 or G = 24 (depending on whether or not F = k). Both A 4 and S 4 have no element of order 6, and these are the only groups of order 12 and 24, respectively, with this property. One can see this by simply checking all isomorphism types of groups of order 12 and 24. (For a convenient list of the groups of order 24, see Appendix C in [17].) Therefore it will suffice to show that in either case G has no element of order 6. If φ Gal(L/k) and φ K has order 1 or 2, it is easy to see that φ has order 1, 2, or 4. If φ K has order 3, straightforward computations will show that φ must have order 3. (One need only check the order of the action on β 12 and β 13.) For our proof of Lemma 1 we will need some basic properties of the absolute logarithmic Weil height (or simply height ) of algebraic numbers. The definition of the height can be found in [2], along with all of the properties stated below. We denote the height of an algebraic number α by h(α). We will use four basic properties:

6 6 ON THE COMPOSITUM (i) h(α) = 0 if and only if α is a root of unity, (ii) h( α) = h(α) for any algebraic number α, (iii) for any algebraic number α, rational number λ, and any choice of the root α λ, we have that h(α λ ) = λ h(α) (in particular, h( α) = 1 2h(α)), and (iv) if K is any number field, every nonempty subset of K contains an element of minimal positive height. The fourth property follows from Northcott s Theorem, which states that the set of algebraic numbers with height at most T and degree at most d is finite for any positive T and d. Proof of Lemma 1. We will actually find a generator of K/F of norm 1. Let E be the set of all α in K such that N K/F (α) = 1, α is not a root of unity (i.e. h(α) > 0), and α generates K/F (i.e. α F ). First, we claim that E is not empty. Without loss of generality, we may write (8) f(x) = x 3 + px + q with p and q in k, q 0. Let ξ 1, ξ 2, and ξ 3 be the roots of f(x). Notice that for any r k the roots ξ i r of the shifted polynomial f r (x) = f(x + r) are also generators of K/F. For each r k, define ( (9) g r (x) = x ξ )( 1 r x ξ )( 2 r x ξ ) 3 r. ξ 2 r ξ 3 r ξ 1 r It is straightforward to verify that g r (x) k[x], and a routine calculation shows that the discriminant of g r (x) is (10) r = 27qr3 + 9p 2 r 2 + p 3 (r 3 + pr + q) 2. For each r k, let η r = ξ 1 r ξ 2 r and notice that N K/F (η r ) = 1. We have that Gal(K/F ) acts transitively on {ξ 1, ξ 2, ξ 3 }, and hence on {η 1, η 2, η 3 }. Notice that there are only finitely many roots of unity in K, and only finitely many values of r for which r = 0. Therefore, since η r takes on infinitely many values as r varies in k, we may choose a value of r such that r 0, and η r is not a root of unity. Since r 0, the roots of g r (x) are distinct, so η r is not fixed by Gal(K/F ). This implies that η r F, so η r E, proving our claim that E is not empty. By property (iv) of the height, there is some α 1 E of minimal height. Notice that α 1 is not a square in K if it were, then either α 1 or α 1 would be an element of E of strictly smaller height than α 1, by properties (i), (ii), and (iii). Let α 2 and α 3 be the Galois conjugates of α 1 over k. Since α 1 has norm 1, we have that α 1 α 2 = α3 1 is also not a square. (This is a special case of the fact that a generator of a cubic extension K/F with square norm is itself a square in K if and only if a product of the generator and a conjugate is a square in K.)

7 ON THE COMPOSITUM 7 3. Proof of Theorem 1 By Proposition 1, parts (a) and (b) follow from the solutions to the embedding problems in Propositions 2 and 3. That is, every irreducible quadratic splits in the splitting field of some irreducible cubic, and every irreducible quadratic or cubic splits in the splitting field of some irreducible quartic. The following lemma will be used to prove part (c). Lemma 2. If d is an integer greater than or equal to 5, then there exists a prime number p ( d 2, d) such that if G is a transitive subgroup of S d containing a p-cycle, then either G = S d or G = A d. Proof. The transitive groups of degree d are well-known for small d see for example [4] for the groups up to degree 11; GAP (see [13], [14]) has a library of all of them for d 30). It can be checked easily that we can use p = 3 when d = 5, and we can use p = 5 when d = 6, 7; in each of these cases, S d and A d are the only transitive subgroups with order divisible by p. Therefore all that remains is to prove our lemma in the case d 8. There exists at least one prime p ( d 2, d 2). This can be verified easily for small d, and for large d it follows from any result slightly stronger than Betrand s postulate, which states that there is always a prime in (m, 2m 2) for m > 3 (proved by Chebyshev). For example, M. El Bachraoui shows in [11] that there is always a prime in [2m, 3m] for m > 1. Let G be a transitive subgroup of S d containing some p-cycle g. Without loss of generality, g = (1 2 3 p). Since G is transitive, for each i {p + 1,..., d} there is some element σ i G such that σ i (1) = i. If we let g i = σ i gσi 1, then g i will be a p-cycle in G whose support contains i. Since p is prime, each g i acts primitively on its support, which is a set of size p. Since p > d 2, the pairwise intersections of the supports of the groups g i are nontrivial. Therefore we can apply Proposition 8.5 from [19] inductively to see that the subgroup H = g, g p+1, g p+2,..., g d is a primitive subgroup of S d. Since H contains a p-cycle and p < d 2, Theorem 13.9 from [19] tells us that either H = S d or H = A d, and since H G, our proof is complete. Part (c) will be an immediate corollary of the following theorem. Theorem 5. Let G sd G 1 G n be a subdirect product of transitive groups of degree d, where d 5. Then there is some prime p < d such that the cyclic group C p is not a quotient of G. Proof. Fix d 5. By Lemma 2, there is a prime p ( d 2, d) such that the only transitive subgroups of S d containing a p-cycle are S d and A d. We proceed by induction on n, noting that the case n = 1 follows immediately by our choice of p. In general, we will have that G sd G 0 G n, where G 0 is a subdirect product of n 1 transitive groups of degree d. If N is any normal subgroup of G, we have that N sd N 0 N n for some normal subgroups N 0 G 0 and N n G n. We may write G as a fibered product G = G 0 Q G n, i.e. G 0 and G n respectively admit epimorphisms α 0 and α n onto some group Q, and G = {(a, b) G 0 G n α 0 (a) = α n (b)}. Similarly, we have N = N 0 R N n for some group R which is a quotient of both N 0 and N n. By the inductive hypothesis, neither G 0 /N 0 nor G n /N n has order p. Suppose that G/N = C p. Since G/N surjects onto both G 0 /N 0 and G n /N n, the latter two groups must be trivial.

8 8 ON THE COMPOSITUM Therefore (11) p = G N = G 0 G n / Q N 0 N n / R = G 0/N 0 G n /N n R Q = R Q. By inspection of the possibilities for Q and R, (11) cannot hold. This establishes part (c) of Theorem 1. Indeed, it shows that k [p] k [d], for p and d as above, whenever k is any field that admits a degree p Galois extension. We conclude this section by demonstrating that part (a) of Theorem 1 can be proved by elementary methods when k = Q. Elementary proof that Q [2] Q [d] for all d 2. It will suffice to show that p Q [l] for any prime l, whenever p is a rational prime or p = 1. For p prime or equal to ±1, set (12) f p (x) = x l l(lp + 1)x + (l 1)(lp + 1) The discriminant p of this polynomial is given by the following (see for example [?]): (13) ( 1) (l 1)(l 2)/2 p = (l 1) l 1 l l+1 (lp) l 1 p. In particular, p will be in the splitting field of either f p (x) or f p (x). One can check the irreducibility of the polynomials f p using the Eisenstein-Dumas criterion applied to shifts of the polynomial. Notice that ramification of primes makes a generalization of this simple proof to number fields unreasonable; of course Neumann s theorem mentioned above already gives this result in a stronger way every quadratic splits in the splitting field of a single degree d polynomial. We begin with a lemma. 4. Unboundedness: proofs of Theorems 2 and 4 Lemma 3. Let p be an odd prime number, and let (14) G = D n 1 p C p = r 1, s 1,..., r n 1, s n 1, r n be the direct product of n 1 copies of the dihedral group D p and a cyclic group of order p, where for, i {1,..., n 1}, the i th D p = r i, s i is generated by the p-cycle r i and the 2-cycle s i, and C p = r n. Let (15) H = r 1 r n, r 2 r n,..., r n 1 r n G. If B is a subgroup of G with H B G, then r n B. In particular, the intersection of all such subgroups B strictly contains H. Proof. Let G p = r 1,..., r n be the unique Sylow p-subgroup of G, considered as an n- dimensional F p -vector space. Any Sylow 2-subgroup G 2 will be an (n 1)-dimensional F 2 -vector space which acts by conjugation on G p, so that G = G p G 2. Let H B G. Note that H is a codimension 1 subspace of G p, so if B contains any element of order p not in H, then B contains all of G p. If B contains any involution τ G, notice that there will be some i such that τ acts non-trivially on the i th copy of D p, so that r i r n, τ will contain r n. Since every nontrivial element of G is either of order p, an involution, or of order 2p (a power of which is an involution), this completes our proof.

9 ON THE COMPOSITUM 9 Corollary 2. Let k be a number field or global function field, and let p be an odd prime number. Then k [p] /k is unbounded. Proof. Let G and H be as in Lemma 3. Since G is solvable we have an extension L/k with Gal(L/k) = G. Let L H be the fixed field in L of H. It is clear from our construction that [L H : k] = p 2 n 1. The Galois correspondence tells us that every proper subextension of L H /k corresponds to a subgroup B of G with H B G. Furthermore, since the intersection of all such groups strictly contains H, the compositum of all proper subextensions of L H /k is strictly a subfield of L H. This shows that L H is not generated by elements of degree less than p 2 n 1. Notice that the field extension L H /k in the proof above is not Galois (H is not normal in G). As we will prove in the next section, this was necessarily so. We write H p for the finite Heisenberg group of order p 3, when p is a prime. This group is defined as the multiplicative group of upper triangular matrices of the form 1 a c 0 1 b with a, b, and c belonging to the finite field F p. The group H p plays an important role in our Galois unboundedness results. We review some of its properties. First, H p has a natural action on a three-dimensional vector space over F p. Analyzing this action, it is easy to see that when an element of H p acts on a vector, the third coordinate is fixed, and H p acts faithfully and transitively on a 2-dimensional affine subspace, which has p 2 elements. Thus we see that H p is isomorphic to a transitive subgroup of S p 2. The group H p is an extraspecial p-group, meaning its center, commutator, and Frattini subgroups coincide and have order p. For our purposes the only relevant fact is that the commutator subgroup is cyclic of order p. We can construct larger extraspecial p-groups as follows. Let n be a positive integer, and inside of the direct product Hp n consider the normal subgroup (16) N p,n = {(z a 1 1,..., zan n ) Σ n i=1 a i 0 (mod p)}, where z i generates the center of the i th copy of H p. The quotient H n p /N is an extraspecial p-group of order p 2n+1 and exponent p, which we will denote by E p,n. The following lemma can be found in [5], and follows from some basic facts about the extraspecial p-groups (see for example [1] and [10]). Lemma 4. The intersection of all subgroups of index less than p n in E p,n contains the commutator subgroup. In particular, this intersection is nontrivial. S. Checcoli and U. Zannier used this fact in [7] to show that, for a number field k, the extension k (d) /k is not in general Galois bounded. (These ideas later appeared as part of a more comprehensive treatment in [5].) The idea of using these groups is attributed to A. Lucchini. However, the authors were not concerned with the question of which values of d suffered from this pathology, nor with the more general question of the boundedness of k [d] /k. The use of extraspecial p-groups (which are certainly not the only groups with properties like the conclusion of Lemma 4, but are natural and easy to work with) remains,

10 10 ON THE COMPOSITUM our primary tool for proving that extensions are Galois unbounded. The following lemma simplifies our application of this principle. Lemma 5. Let d be a positive integer. Suppose there is a prime number p such that there is a solvable group G which is a subdirect product of transitive groups of degree d, and a quotient of G is isomorphic to H p. Then k [d] /k is Galois unbounded for any number field or global function field k. Proof. By Shafarevich s Theorem, for any positive integer n we can realize G n as the Galois group of some extension L/k, and we will have L k [d]. There will be Galois subextension K/k with Galois group Hp n, and the subfield of K corresponding to the normal subgroup defined in (16) will have Galois group E p,n, and will therefore not be generated by elements of degree less than p n. Proof of Theorem 4. Part (a) follows immediately from Lemma 5, since H p is solvable and transitive of degree p 2. Checcoli showed how to realize these groups explicitly in [5]. Let d = 2m, and consider the partition {1,..., d} = {1,..., m} {m + 1,..., 2m}. Let G 1 be the dihedral group (17) G 1 = r 1, s 1 r m 1 = s 2 1 = 1, r 1 s 1 = s 1 r 1 1 of order 2m acting on {1,..., m}, and let (18) G 2 = r 2, s 2 r m 2 = s 2 2 = 1, r 2 s 2 = s 2 r 1 2 be an isomorphic copy of the same group, acting on {m+1,..., 2m}. Let H be the subgroup of S d generated by the involution (1 (m+1)) (2 (m+2)) (m 2m), which interchanges the blocks of the partition. Notice that this involution normalizes G 1 G 2, so we may consider the semidirect product (19) G := (G 1 G 2 ) H. Let N = r 1, r 2 be the normal subgroup of G generated by the rotations. The quotient G/N is isomorphic to the dihedral group D 4, which is isomorphic to the Heisenberg group H 2 of order 8. Applying Lemma 5 again gives part (b). To prove part (c), we observe that, if p is a prime and m 1 (mod p), then there is a transitive group of degree pm of the form Cm p H p. We now describe how to construct this group, whose action on {1,..., pm} respects the partition into p blocks of size m. Let σ i be an m-cycle permuting the elements in the i th block, for i {1,..., p}. Each of these m-cycles is normalized by some (m 1)-cycle. Since m 1 (mod p), some power of this (m 1)-cycle will be a product of disjoint p-cycles; we denote this product of p-cycles by τ i. We set α = τ 1 τ 2 τ p, β = τ 2 τ 3 τ p = τ1 1 α, and let γ be a permutation of order p which permutes the blocks and normalizes all of the above permutations. The permutations σ i generate a group isomorphic to Cm, p while α, β, γ is isomorphic to the Heisenberg group H p. The desired group is then σ 1,..., σ p, α, β, γ. It would be quite tedious to write explicitly the generators of the group constructed in the proof of part (c) for general p and m, but we will make this construction more clear by giving an example with d = 21 = 3 7. The 7-cycle σ 1 = ( ) is normalized by the 6-cycle ( ). Squaring this permutation yields a product of 3-cycles τ 1 = (2 5 3)(6 7 4),

11 ON THE COMPOSITUM 11 which normalizes σ 1. Proceeding in the same way with the other two blocks of {1,..., 21}, we will obtain the following generators: σ 1 = ( ); σ 2 = ( ); σ 3 = ( ); α = (2 5 3)(6 7 4) ( )( ) ( )( ); β = ( )( ) ( )( ); γ = (1 8 15)(2 9 16)( )( )( )( )( ). The interested reader will verify that this group is as described and see clearly that the construction for general p and m will work in much the same way. Notice that α, β, γ acts on the two nine-element sets {2, 3, 5, 9, 10, 12, 16, 17, 19} and {4, 6, 7, 11, 13, 14, 18, 20, 21} in unison, and that this action on nine points is the same as the natural action of the Heisenberg group on the 2-dimensional affine space described above. We had to choose this embedding of the H 3 into S 21 so that it would normalize our 7-cycles. Proof of Theorem 2. We know that k [2] = k (2) ab, so k[2] /k is bounded. If d > 2, then d is divisible by c, where c is either 4 or an odd prime. By Corollary 2 and part (b) of Theorem 4, k [c] is unbounded over k, and k [c] k [d]. We remark that our proofs actually demonstrate that k [d] /k is also unbounded in the case where k is a global function field and d Galois boundedness in prime degree In this section we prove Theorem 3. Clearly the general technique for showing boundedness is to find subgroups of small index inside of a Galois group G, whose intersection is a given subgroup H. If we want to show Galois boundedness, we take H to be normal. We will show that we can accomplish this task when G is a subdirect product of transitive groups of prime degree. The following lemma characterizes the transitive groups of degree p. Lemma 6. If p is a prime number and G is a transitive group of degree p, then we have G = T B, where T is simple and B is a subgroup of C p 1. This lemma can be proved by elementary means, as was pointed out by an anonymous referee. It can also be seen quickly using the classification of finite simple groups: a theorem of Burnside (see [9], Theorem 4.1B; cf. [19], Theorem 11.7) implies that G is either a subgroup of C p C p 1 containing C p, or an almost simple group, meaning that there is a simple group T such that T G Aut(T ). The Classification Theorem for Finite Simple Groups implies that there is a very small list of possibilities for T (see [12], corollary 4.2), and the lemma can be easily checked in these cases. We are now ready to establish a group theoretic result, of which Theorem 3 will be an immediate corollary. Theorem 6. Let p be a prime number and let G be a finite subdirect product of transitive groups of degree p. If N is a normal subgroup of G, then N is an intersection of subgroups of index at most p in G.

12 12 ON THE COMPOSITUM Proof. Let G sd G 1 G n, where G i is a transitive group of degree p for i {1,..., n}. If we consider each group G i acting transitively on a set Ω i of size p, we have G acting faithfully on the union of these sets, which we denote by Ω. Let π i denote the projection onto G i. We proceed by induction on n, the case n = 1 following easily from Lemma 6. For each i, we have that G G i is a normal subgroup of G. If G G i is trivial for any i, then we may appeal to the inductive hypothesis, for G acts faithfully on Ω \ Ω i. Therefore, for each i we will assume that, G G i is a nontrivial normal subgroup of both G and G i, and so G G i contains the minimal normal subgroup T i of G i. Let T = i T i; by Lemma 6, we have that G/ i T i is abelian of exponent dividing p 1. By the inductive hypothesis, we can handle the case where N acts trivially on some Ω i as follows. We will consider the action of G on Ω \ Ω i. Let K be the Kernel of this action, and we may write NK of this action as an intersection of some collection of subgroups {H j } j of index at most p in G, using the inductive hypothesis. If N acts trivially on Ω i, then the intersection of these subgroups H j, together with the stabilizers of points of Ω i, will be equal to N, completing the proof in this case. Now we may assume N acts nontrivially on each Ω i, which means that T i π i (N). For each i such that T i is non-abelian (recall that T i is simple), we will have T i = [T i, N] N. Now we see that that G/T N is abelian of exponent dividing p 1, and T N/N is an elementary abelian p-group. We have a short exact sequence 1 T N/N G/N G/T N 1, which splits by the Schur-Zassenhaus Theorem. Let V = T N/N and B = G/T N, so we have (20) G/N = V B. We want to show that there is a collection of subgroups of index at most p in G/N whose intersection is trivial. It is clear that we can find such subgroups whose intersection is V, since B is abelian of exponent dividing p 1. Therefore it suffices to find subgroups of G/N of index at most p whose intersection meets V trivially. Considering the F p - vector space V as a B-module, Maschke s Theorem tells us that V decomposes as a direct sum of irreducible B-modules. Since x p 1 1 splits over F p, it follows that these irreducible submodules are one dimensional. Now we have submodules V i of index p (codimension-one submodules), which yield subgroups V i B of index p in G/N, and the intersection of all of these meets V trivially. Combining this with the groups mentioned in the previous paragraph, we have a collection of subgroups of index at most p in G/N having trivial intersection. Proof of Theorem 3. Let k be any field, and let K/k be a finite Galois subextension of k [p] /k, where p is prime. This implies that K is contained in a degree p compositum L/k. Let G = Gal(L/k) and N = Gal(L/K). Then G isomorphic to a subdirect product of transitive groups of degree p, and N is normal in G. Theorem 6 implies that N is an intersection of subgroups of index at most p in G. By the Galois correspondence, this means that K is the compositum of finitely many extensions of k of degree at most p. Therefore, K/k is generated by elements of degree at most p. (In fact, it must be generated by elements whose degrees are either equal to p or divide p 1.)

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